Air transport system

ABSTRACT

A pneumatic elevator for containers is provided wherein a vertical duct has end portions with nozzles in the form of holes extending through the ends and entering the duct in a downstream direction. The sides of the duct are provided with guide rails and at least one side is provided with a plurality of apertures along the length thereof through which air, introduced into the duct through the end nozzles, is exhausted. The size and angle of the nozzles are correlated with the static pressure differences across the jets to provide the requisite duct static pressure with respect to the size and number of apertures in the inner side which determines the volume of air exhausting from the can duct. Differential static air pressure separates each can from the other to minimize scratching and denting. The lower end of the elevator may be provided with an in-feeder to feed cans into the elevator from storage or from an accumulator and the upper end of the elevator may exhaust into a deaccelerator which slows down the cans coming from the elevator and positions them at the proper speed and spacing for the subsequent work station.

FIELD OF INVENTION

This invention relates to a container transportation system in which airis the transport medium and the containers provide dynamic reaction withthe air and coact with the transport structure to control and transportthe containers. The transport system may include an in-feeder toestablish a predetermined spacing between containers, an elevator formoving containers between two different levels at the predeterminedspacing or other spacing and a de-accelerator for de-accelerating thecontainers for introduction into a work station or a combination of anyor all of these. Additionally, the system may include a fan conveyor topermit delivery of containers in an upright position as well as ahorizontal conveyor of containers in either upright or horizontalposition.

DESCRIPTION OF PRIOR ART

Air lift conveyors have been used for conveying various kinds ofparticulate material as well as packages or similar articles. Adisadvantage of the prior art lift conveyors is that the lift andtransport is dependent upon the shape and arrangement of the air nozzleswhich differ with the material being handled. Usually the nozzles are inthe form of holes extending through the deck or side wall of a conveyorand entering the conveyor channel in a downstream direction. Anotherpopular form resembles a series of louvers, U.S. Pat. No. 3,759,579,providing laterally extending slots through which the air passes with asubstantial downstream velocity to not only support the material butalso to propel it along the conveyor channel.

Most elevators, that is conveyors for vertical transport of articles andin particular cans, make use of a belt arrangement where belts areprovided on each side of the cans to grip the cans and propel them in avertical direction. The belts on either side of the cans travel in adirection such that the belts in contact with the cans move in the samedirection.

If the tension on the belts is too great, the can per minute rate (CPM)of the elevator will be reduced because cans cannot be fed fast enoughand may be defomred. If the belt tension is too loose, the belts willslip on the cans and the elevator will jam. Additionally, if the beltsslip on the cans they may be severely marked.

The jamming of the mechanical elevators is a major problem and time lostin downtime to clear such jams is both costly and dangerous. Thefrequent tuning and adjusting of the belts is necessary to assure evenreasonably satisfactory operation. This requires that the elevator beshut down for a period of time. Many mechanical elevators are as high as40 or 50 feet and belt tensioners are positioned on both sides atintervals from 12 to 16 inches. In most cases, three belts are used, twoon one side and one on the other, all of which must be adjusted. Whenthe belts stretch, portions must be removed to compensate. With losttime production rates in the can industry in the range of $16,000 to$20,000 an hour and an average clearing time of a jam in an elevatortaking two to five minutes, with jams occurring on an average of two anhour, the monetary loss in a 24 hour plant is considerable.

The safety problems with mechanical elevators are quite extensive. Thecan per minute (CPM) rate of mechanical elevators is restricted as thespacing between cans is usually about one foot. Thus, a 1000 CPM raterequires a 1000 feet per minute (FPM) operation of the elevator. Ashigher speed equipment is introduced, elevator speeds must be increasedto maintain production. As faster and faster speeds are required, therate of travel of the belts that grip the cans reaches a high enoughvelocity that the slippage on the can at the pick-up point isself-defeating. There are also problems experienced with reverserotation of the cans at discharge with these high can speeds. Further,the danger of can damage becomes greater due to can-to-can contact aswell as possible destruction of the machinery at such high speeds.

U.S. Pat. No. 4,010,981 is directed to an "Air Conveyor With TunnelGuide" which includes an air conveyor combined with a tunnel guidemember. The air conveyor includes an elongated plenum chamber having aperforated deck having transverse slots forming directional air jets. Animperforate wall tunnel guide member confronts the air jets and extendsalong the conveyor to define a conveying zone. The guide has for itspurpose the trapping and conserving of the conveying air and air gapsare provided between the marginal edges of the tunnel guide wall and theconveyor surface to prevent build up of undesired back pressure orstatic pressure within the conveying zone.

SUMMARY OF THE INVENTION

According to the invention, a can duct is provided having end portionswith nozzles in the form of holes extending through the ends andentering the can duct in a downstream direction. The sides of the canduct are provided with guide rails and at least one side, such as theside forming the inside radii, is provided with a plurality of aperturesalong the length thereof through which air introduced into the can duct,through the end nozzles, is exhausted. The size and angle of the nozzlesare correlated with the static pressure (SP) difference across the jetsto provide the requisite can duct static pressure with respect to thesize and number of apertures in the inner side which determine thevolume of air exhausting from the can duct. The nozzles in the ends ofthe can duct are supplied with air from air supply ducts of which thecan duct ends may be one wall and the supply ducts extend for the fulllength of the can duct; however, the air supply ducts may be made up ofa plurality of shorter ducts supplied by a number of fans or a fulllength duct with a single fan.

The lower end of the elevator may be provided with an in-feeder to feedcans into the elevator from storage or an accumulator such as a singlefiler of the type set forth and disclosed in my application for U.S.patent Ser. No. 869,371, filed Jan. 13, 1978. The upper end of theelevator may exhaust into a de-accelerator which slows down the canscoming from the elevator and positions them in speed and spacing to bereceived by subsequent work stations and the like.

Referring now to the drawings:

FIG. 1 is a front view in elevation of a portion of the transportationsystem according to this invention including an in-feeder section, anelevator section and a de-accelerator section partly in cross sectionand partly broken away to conserve space;

FIG. 2 is a side elevational view in cross section and partly brokenaway;

FIG. 3 is a cross sectional view taken along line 3--3 of FIG. 2;

FIG. 4 is a cross sectional view along line 4--4 of FIG. 3;

FIG. 5 is a diagrammatic representation of the de-accelerator section;

FIG. 6 is a view in cross section of a variable angle nozzle embodiment;

FIG. 7 is a cross sectional view of an air elevator fan section similarto that shown by dotted lines in FIG. 1 as well as a horizontal sectionwith parts broken away to conserve space;

FIG. 8 is a diagrammatic representation of the cross sectional portionof FIG. 1 illustrating the velocity components of the air and can;

FIG. 9 is a graphical representation of the Power Transfer to Can Due toJet Airstream and Can Velocities;

FIG. 10 is a graphical representation of the Force of Air on Can Due toJet Airstream and Can Velocities;

FIG. 11 is a graphical representation of the Jet Diameter Constant(K_(j)) for different jet diameters;

FIG. 12 is a graphical representation of the Vent Diameter Constant(K_(v)) for different vent diameters;

FIG. 13 is a graphical representation of the Air Jet Velocity vs JetDiameters; and,

FIG. 14 is a schematic representation of the forces acting on a canwhich are weight, w, force due to the air jets in a can duct, F₁ ' andF₂ ', and force due to pressure difference across the can (P₁ -P₂) A.

Referring now to FIGS. 1, 2, 3, and 4 of the drawings, there is provideda vertically positioned elevator referred to generally by the referencenumeral 10 and extending roughly from point B to point C. The embodimentdisclosed is designed for use in elevating empty cans 12 which have oneend 14 open and one end 16 closed. The elevator body 11 is generally ofrectangular cross section providing a rectangular can duct 18corresponding substantially to the profile of the cylindrical cans 12.The can duct size is controlled close enough to the actual can size thatan effective workable air seal is established around each can in theduct, while permitting easy passage of the cans through the duct. Thecan duct has end guides 20 and 22 adjacent the open end and the closedend of the cans 12. The end guides also form one wall of rectangular airsupply ducts 24 and 26 which are connected to a suitable supply of airsuch as from fans as at 25 and 27. The can duct is also provided with apair of sides 28 and 30 of a length just slightly greater than thelength of the can sufficient to provide clearance between the end guidesand the can. A pair of guide rails 32, 34 and 36, 38 are positioned ingrooves 40, 42 and 44, 46 in the sides 28 and 30, respectively, adjacenteach edge thereof and positioned adjacent the end guides 20 and 22. Theguide rails are positioned to extend only slightly above the innersurface of the sides 28 and 30 of the can duct 18 to provide a verysmall clearance 48 and 50 between inner surface of the sides 28 and 30and the sides of the can 12.

A series of air jet holes 51 and 52 are provided in end guides 20 and 22along the length thereof in approximately centered relationinterconnecting the air supply ducts 24 and 26 with the can duct 18 ateach end thereof. The holes 51 and 52 are bored at an angle to enter thecan duct 18 in a downstream direction and are positioned at an angle inthe range from about 18° to about 60° off normal to the longitudinalaxis of the can duct. Where the end guides 20 and 22 are of insufficientthickness to provide the necessary directivity to the bore holes 51 and52, the bores 51a and 52a, FIG. 8, may be provided in jet boards 54 and55 of sufficient thickness to provide the desired directivity secured bysuitable means to the wall forming the end guides 20 and 22 common tothe air supply ducts 24 and 26. The jets 51 and 52 may be of differentsizes and angles and of differing sizes and angles depending upon theresults desired as will be explained presently.

At least one of the sides 30 of the can duct 18 is provided with aseries of exhaust apertures 31 serially positioned along the length ofthe can duct communicating the can duct with the immediately adjacentambient atmosphere. The size and the spacing of the holes can be variedto suit a particular set of conditions although the apertures arenormally only located in one of the sides and that preferably being theside forming the inside radius of any curvature such as the dischargeelbow in the elevator for reasons to be explained. The particular sideof the elevator on which the exhaust apertures are located willdetermine the direction of rotation, if any, that will be imparted tothe cans in passing through the can duct. The cans will tend to hug therails positioned on the side containing the exhaust apertures and thusthe rotation of the leading edge of the can will be toward the exhaustopenings. This means that the cans will be in contact or near contactwith those rails on the inside radius of a curve and will avoid reverserotational effects of contact with the rails on the outside radius of acurve. This rotational feature to be preferred as the coefficient offriction is substantially reduced and the marking of cans is minimizedor eliminated.

Also, the rotation effect assures that where the interior of a can hasbeen coated, that the coating remains in suspension until set up. Therotation of the cans with respect to the discharge direction iscontrolled so that the desired rotation is established prior todischarge to avoid a reverse rotational or a non-rotational lag time.This is extremely important in those cases where a high can per minuterate is desired and the discharge velocity of the cans must be held orreduced to a low figure. Where a high can per minute rate is desired,the cans must be spaced close together.

With the air elevator according to this invention, the obtaining of canspacing of four to five inches and closer between cans is achieved andmaintained easily. Thus, the air elevator operating at can speeds of 750to 800 feet per minute (FPM) can deliver 1,400 cans per minute (CPM).Closer spacing can be maintained at the same or lower FPM rates forhigher production rates without the attendant difficulties of themechanical elevator. Should downstream equipment fail and result in cansbacking up, a mechanical elevator must be stopped or at least thefeeding of cans to the elevator must be stopped while the air elevatoraccording to the present invention can fill with cans without danger ofdamage to the cans. Further, the elevator will provide accumulation andthe possibility of jamming is non-existent since pressures that areapplied on the cans are substantially uniform and not at any one point.The present system has the further advantage in that the elevator willclear and resume production without further adjustment.

The air elevator of this invention is a semi-enclosed system withrespect to the can duct. The air velocity from the jets 51 and 52 ispreferably established on the basis of the optimum jet size for theacceptable jet and duct velocity for the optimum power transfer and thelowest cubic feet per minute (CFM) from fans 25 and 27 necessary tosupport and transport a particular can size and weight. The appropriatejet air velocity is established by a static pressure difference acrossthe jets at selected points along its length. From test results basedupon operating systems, certain empirical relationships have evolved.

It has been found that air jet nozzles formed by a 1/4 inch diameterbore will provide a jet air velocity of about 3400 feet per minute (FPM)at a static pressure of 1 inch of water and will require 1.16 cubic feetof air per minute (CFM) according to Equation 1:

    A.sub.j ×V.sub.j =Q                                  (1)

Test results have shown that as the jet nozzle hole size is reducedbelow about 1/4 inch, the velocity is reduced along with the volume flowrate. Where hole sizes larger than 1/4 inch diameter are used, the airvelocity achieved is only slightly greater but the cubic feet per minuteof air delivered is considerably greater. The air elevator of thisinvention is predicated on optimizing the design to operate at thelowest static pressure that will give an efficient jet air velocity withrespect to the required can velocity, and at a jet diameter that willprovide sufficient C.F.M. to give the necessary lift to maintain therequired can flow rate.

For best results and lowest friction loss through the holes or nozzles51 and 52, the wall thickness or jet boards 54 and 55 through which thehole is bored is preferably not thicker than one and one-half thediameter of the hole nor so thin in relation to the hole direction as topermit any significant amount of air to pass through the hole withouthaving a downstream directivity imparted thereto.

The angle of the holes or bores 51a and 52a, FIG. 8, through the jetboards 54 and 55 affects the speed of the cans in the can duct and foran elevator operating at a can speed of about 750 to 800 FPM to handleabout 1400 cans per minute, the holes are preferably positioned at about45° with respect to a line normal to the direction of can flow for thedesired operation. The air issuing from the jets 51a and 52a impinge onthe can and transmit a force 53 which can be broken down into componentsor vectors including a vertical force represented by arrow 56 as well asa horizontal force represented by arrow 58, FIG. 4. The forcesrepresented by arrow 58 tend to center the can endwise in the can ductand the forces represented by arrow 56 tend to move the cans up orthrough the can duct.

The definition of terms with units used in the equations are set forthin Table I.

TABLE I GLOSSARY

P_(o) =Power (lbf-ft)/min

F=Force (lbf)

M=Mass flow rate of cans (lbf-min)/ft.

m=Mass flow rate of air (lbf-min)/ft.

V=Velocity of cans (ft/min)

V_(j) =Velocity of air through jet (ft/min)

V_(v) =Velocity of air through vents (ft/min)

V_(r) =Relative velocity of air with respect to can

W=Weight of all cans in elevator at one time (lbf)

g=Acceleration of gravity (ft/min²)

t=Time (min)

P=Static pressure in can duct (in.H₂ O)

φ=Static pressure in supply duct (in.H₂ O)

Δφ=Static pressure difference across jets between supply and cand duct(in.H₂ O)

A=Cross sectional area of can (ft²) through the axis of rotation

K=Velocity constant (ratio of can to jet velocity) (Dimensionless)

θ=Angle of jet from horizontal (Degrees)

D_(j) =Diameter of jet (Inches)

D_(v) =Diameter of vent (Inches)

Q=Volume flow rate of air (ft³ /min)

A_(j) =Area of jet (ft²)

A_(v) =Area of vent (ft²)

N=Number of jets per vent (Dimensionless)

ρ=Air Density (lbf-min²)/ft⁴

K_(d) =Diameters constant (Dimensionless)

K_(j) =Jet velocity constant (Dimensionless)

K_(v) =Vent velocity constant (Dimensionless)

K_(f) =Force constant (Dimensionless)

w=Weight of one can.

Since the air jets 51a and 52a are introducing air at intervals of from3/4 to 11/2 inches along the ends of the can duct for its entire length,the cubic feet per minute of air introduced will essentially double forany given interval. Air must be allowed to be discharged through theexhaust holes 31 located in at least one of the sides of the can duct.If air is not allowed to escape at a controlled rate along the length ofthe can duct, cans will not be allowed to enter the elevator or beintroduced into the can duct area. On the other hand, if too much air isallowed to escape, the cushion of air between cans, which is a functionof static pressure, will not be maintained and one can will thenovertake the next can and, with each can depriving the other of theavailable energy, the cans will back up in the elevator. The vent andjet sizes are chosen to establish a pressure ratio across each can inthe can duct, see Equation 16. The pressure is slightly less above eachcan and slightly greater below each can on a gradient all the way up theelevator, the greatest pressures being at the infeed decreasing all theway to ambient at the discharge. The pressure gradient offsets apredetermined portion of the weight of each can by giving the can anadded lift by acting on the longitudinal cross sectional area of thecan. The "effective weight" of each can is that portion of the weight ofeach can that is balanced by the lift due to the static pressuredifference across each can in the can duct. Thus, it is seen that thevolume of air introduced with respect to time in combination with thejet velocity as well as the duct velocity and the amount of airexhausted, and where, determines the can spacing and maintains thespacing established at the feed end of the elevator. The elevator andthe feed must be adjusted and balanced for optimum results. It will beappreciated that the openings 31 may be holes of various shape, or acontinuous elongated slot in one or more of the sides running the lengthof the can duct.

The principle of the air elevator operation according to this inventionis that can movement up the elevator is accomplished efficiently byusing a balance of two sources of power. The first of these powersources is the air jets positioned in the end guides of the elevator,which are used to transfer energy from the mass flow rate of the airthrough the jets to the can ends. This power source provides the forcenecessary to maintain the momentum of the can, i.e. maintain itsvelocity at the required can per minute rate, and also to provide thatportion of the total force required to offset the weight of the cans orto provide the lifting force needed to support the can column. Thissource then provides all of the kinetic energy at work on the system anda portion of the potential energy required to lift the column of cans.The equation defining this force is:

    F=ρA.sub.j (V.sub.j Sin θ-V).sup.2               (4)

This equation is derived from an analysis of the air flow through thejets 51a and 52a depicted diagrammatically in FIG. 8.

Since

    F=MA(Force=Mass times Acceleration)                        (5)

    F=mV.sub.r (Force=Mass Flow Rate times Velocity)           (6)

Where:

V_(r) =the relative velocity of the air with respect to the can in thedirection of the can motion. So

    V.sub.r =V.sub.j Sin θ-V                             (7)

The mass flow rate is

    m=ρA.sub.j V.sub.r                                     (8)

    F=(ρA.sub.j V.sub.r)V.sub.r =ρA.sub.j V.sub.r.sup.2 (9)

so

    F=ρA.sub.j (V.sub.j Sin θ-V).sup.2               (10)

The second source of power is that of the infeeder 60 (whether it be bygravity, mechanical means, or air). The infeeder establishes the abilityto provide a back pressure at the elevator entrance which in turnenables a greater than ambient static pressure to be maintained betweenthe cans.

The back pressure or can duct static pressure provides two positiveeffects in the operation of the elevator; the first is that bymaintaining a greater than ambient can duct static pressure, an aircushion is established between the cans so that when the elevator isbeing fed intermittently and a single can enters and overtakes the pack,its velocity will be retarded by the air it must displace between thecans, thus avoiding impact between the cans which would cause candamage. The second positive effect of maintaining this positive can ductstatic pressure is that it can be used to compensate for the cumulativeweight of cans above any one point in the elevator. For an example,assume that the cans in the elevator are being supported only by thestatic pressure between the cans. The can duct size is controlled closeenough to the actual can size that an effective workable air seal isestablished around each can in the duct, within limits that allow easypassage of the can through the duct. The static pressure between eachcan then will vary according to the number of cans above it, thegreatest can duct static pressure being at the elevator entrance and thelowest pressure at the discharge. Since the supply duct static pressureis constant all the way up the elevator, this can duct static pressuremay be controlled either by maintaining a constant jet diameter whilevarying the outlet vent size or by varying the jet diameter whilemaintaining a constant outlet vent size or a combination of both. Theforce created by the static pressure between each can and acting on thearea of the can is equal to the effective weight of all the cans abovethat point in the elevator. The pressure to area relationship, intheory, is described by the following relationships.

Since the volume flow rate in through the jets is equal to the volumeflow rate out through the vents:

    Qin=Qout, where Q=Volume Flow Rate                         (11)

    Qin=V.sub.j A.sub.j                                        (12)

    V.sub.j =3400√Δφ                          (14)

    V.sub.v =3400√P                                     (15)

and 3400 is in units of feet/minute so

    3400√Δφ NA.sub.j =3400√P A.sub.v ##EQU1##

Thus, by using the effect of both the force through the jets and thestatic pressure variance between the cans, a condition is achievedwhereby the energy needed to raise the can (the potential energy of thesystem) can be supplied by the static pressure between the cans and theenergy needed to maintain the can velocity (the kinetic energy of thesystem) is supplied by the jets. Also any variation of this can beachieved by allowing the jets to offset a portion of the weight therebydetermining the back pressure and the infeed force required.

In analyzing the equation of force through the jets, one can determinethe power transfer from the air to the cans. Since

    F=ρA.sub.j (V.sub.j Sin θ-V).sup.2               (17)

and since Power=Force times Velocity, the power imparted to the can is

    P=ρA.sub.j (V.sub.j Sin θ-V).sup.2 V             (18)

Assuming a constant jet air velocity, the maximum and minimum values ofpower transfer can be found by determining where the slope of the powerequation=0.

Since the slope of the equation is equal to the derivative of theequation ##EQU2## so maximum power transfer occurs where V=1/3V_(j) Sinθ and minimum occurs when V=V_(j) Sin θ as illustrated by FIG. 9.

Analyzing the Force equation

    F=ρA.sub.j (V.sub.j Sin θ-V).sup.2               (17)

and assuming a constant can velocity one can determine that as the jetvelocity increases with respect to the can velocity that the forceacting on the can increases. This force transfer is not linear with thepower transfer since the power depends also on the velocity of the can,not just the relative velocities of the air to the can (see FIGS. 9 and10).

In comparing the graphs of FIGS. 9 and 10, one can see that as thevelocity of the can with respect to the jet air velocity goes above 1/3,the power transfer decreases and the force acting on the can alsodecreases. On the other hand, as the velocity of the can with respect tothe jet air velocity goes below 1/3 the power transfer decreases againbut the force acting on the can increases geometrically. For thisreason, the preferred transporter design will use a jet velocity ofabout 3 times the velocity of the can. However, when transporting thecans in a vertical direction another force enters in, which is theacceleration of gravity, and is seen as the weight of each can. Thisweight must also be balanced, at least in part, by the force of the airthrough the jets. Therefore, a more desirable velocity ratio forvertical transport would be around 7 to 1 where the force of the airacting on each can is about 9 times what it was at 3 to 1 and yet theefficiency of power transfer has dropped to only about 70%. Where theweight of each can increases or where the total can column weightincreases, velocity ratios above 7 to 1 to as high as 25 to 1 may berequired. Thus, a compromise between the force required to lift the cansat the required rate and the efficiency of elevator operation due to thepower transfer must be achieved.

The total pressure of air flowing in a duct is the sum of the staticpressure and the velocity pressure, the static pressure being thebursting pressure (the pressure measured perpendicular to the flow), andthe velocity pressure being the impact pressure (the pressure measuredparallel to the flow). For isentropic (frictionless) flow, the value oftotal pressure can be considered to be a constant. Therefore, if thevelocity of the air decreases (decreasing the velocity pressure), thestatic pressure will immediately increase. Since the flow through thesupply duct is normally below 2000 ft./min. and turbulent, giving a highstatic pressure to velocity pressure ratio, almost all of the air massflow rate through the jets is achieved by means of a high staticpressure differential across the jets. Since the flow of real gases isnot isentropic and there are frictional losses in moving the air alongthe length of the duct, these losses can be overcome by converting thevelocity pressure into static pressure. This is accomplished bydecreasing the velocity of the air in the duct. As the supply duct feedsair into the jets along the length of the duct, the supply duct velocityautomatically decreases because the number of jets requiring airdownstream from each point decreases. The supply duct cross sectionalarea is intentionally held substantially constant throughout the fulllength of the duct to decrease the velocity of the moving air thusincreasing its static pressure. Therefore, with the air supply at thebottom or inlet of the elevator, the static pressure feeding the jetscan be greatest at the outlet even though the total pressure of thesystem has decreased. Thereby the effect of frictional losses in thesystem have been overcome.

The equation relationships are set forth as follows:

Power to move cans at required velocity: ##EQU3##

Force to move cans at required velocity:

    F=MV                                                       (22)

Pressure required in can duct:

    P=W/A                                                      (23)

Velocity required through jets: ##EQU4##

Static pressure difference across jets: ##EQU5##

Static pressure in supply duct:

    φ=P+Δφ                                       (26)

Jet diameter:

(27) From experimental results, it has been determined that a modifieris needed to fit the equation F=ρA_(j) (V_(j) Sin θ-V)² to actualconditions. This modifier K_(f) has been found to equal 1.36approximately. This changes the force equation to F=K_(f) ρA_(j) (V_(j)Sin θ-V)² ##EQU6##

From experimental results it has been found that the velocity of the airthrough the jets is linear and holds to the relationship:

    V.sub.j =3400√Δφ for jets 1/4" or larger in diameter (3)

But for jets less than 1/4" in diameter, it is not linear; and,therefore, the equation must be modified to account for this. Refer toFIG. 11.

Example: Derivation of empirical pressure relationships between supplyduct and can duct. Since volume flow rate out of the can duct (Qout)equals the volume flow rate into the can duct (Qin), the followingrelationships hold:

    Qin=QOut,                                                  (11)

where

    Qin=V.sub.j A.sub.j                                        (12)

    Qout=V.sub.v A.sub.v                                       (13)

    V.sub.j =3400√Δφ                          (14)

(15) V_(v) =3400√P, for holes greater than 1/4" in diameter so 3400√ΔφA_(j) N=3400√P A_(v), where N=number of jets per vent and since Δφ=φ-P,√φ-P A_(j) N=√P A_(v) and (φ-P) A_(j) ² N² =P A_(v) ² ##EQU7## (Same asEquation 16 solved for φ rather than Δφ).

It has been determined that 3 empirical constants are needed to modifythis equation to fit test results. Those constants are K_(d), K_(j) andK_(v). K_(d) is a modifier for the diameters and is described by therelationship ##EQU8## K_(j) and K_(v) are modifiers for the velocity ofthe air through the jet and vent respectively and can be determined fromthe graphs of FIGS. 11 and 12. Tables II through VII set forth theexperimental determination of K_(j) and Tables VIII through XII setforth the experimental determination of K_(v).

So the pressure relationship equation becomes ##EQU9## and solving forA_(v), ##EQU10## And solving for A_(j) ##EQU11##

It must be remembered that in using these relationships they are not alllinear through the full range of variables that can be expected.

It is important to note that the velocity relationship

    V.sub.j =3400√Static Pressure                       (3)

is linear only for jet openings 1/4" or greater in diameter; foropenings less than this, care must be used to determine the correctvalue. The correct value for jets less than 1/4" in diameter is obtainedby multiplying the air velocity through the jet opening at 1" of waterstatic pressure times the square root of the static pressure across thejet opening. Refer to the graphs of FIG. 13. Also, high supply ductvelocities at low static pressure, i.e. 3000 ft/min or greater at lessthan 2" SP, should be avoided. This is an area of non-linearity andcould produce large errors.

The forces acting on the can in the can duct are w, weight of can; F₁ 'and F₂ ', vertical components of force through the jets; and, (P₁ -P₂)A, the force due to pressure difference across the can, where P₁ and P₂are the static pressures in the can duct and A is the cross sectionalarea of the can presented to the can duct.

The forces due to the jets are

    F=mV.sub.r where m=ρV.sub.r A.sub.j                    (6) (8)

so

    F=ρV.sub.r.sup.2 A.sub.j                               (9)

The total force along the y axis due to the jets, FIG. 14, is

    F.sub.T =F.sub.1 '+F.sub.2 '=F.sub.1 Sin θ.sub.1 +F.sub.2 Sin θ(35)

so

    F.sub.T =ρV.sub.r1.sup.2 A.sub.j1 Sin θ.sub.1 +ρV.sub.r2.sup.2 A.sub.j2 Sin θ                 (35A)

since

    V.sub.j =3400√Δφ

    F.sub.T ρ[(3400).sup.2 Δφ.sub.1 A.sub.j1 Sin θ.sub.1 +(3400).sup.2 Δφ.sub.2 A.sub.j2 Sin θ.sub.2 ](35B)

since the jets are normally of the same angle and size at both ends ofthe can, A_(j1) Sin θ₁ =A_(j2) Sin θ₂, thus the equation becomes

    F.sub.T =1.156×10.sup.7 ρ Sin θA.sub.j (Δφ.sub.1 +Δφ.sub.2)                                      (35C)

which represents the force acting along the y axis on one can due to oneset of jets.

Since the experiments referred to in Tables II and III used the force ofthe jets to just support the can weight, the system there described issolved for bodies at rest, e.g. ΣFy=0. Since the can duct 13 is open toambient pressure P₁ and P₂ are approximately equal so the force due tostatic pressure across the can is negligible. Inserting K_(f) as anempirical modifier.

    w=1.156×10.sup.7 K.sub.f ρ Sin θA.sub.j (Δφ.sub.1 +Δφ.sub.2)                                      (36)

                  TABLE II                                                        ______________________________________                                        EXAMPLE 1: Calculation of K.sub.f using experimental results, see             FIG. 14                                                                       ______________________________________                                        w =         can weight = 20.8 Gm = .045864 lbf                                F.sub.1 =   Force applied to open end of can,                                 F.sub.2 =   Force applied to closed end of can                                φ.sub.1 =                                                                             5" H.sub.2 O                                                                             φ.sub.2 =                                                                           5.9" H.sub.2 O                               A.sub.1 =   .00077 ft.sup.2                                                                          A.sub.2 = .00077 ft.sup.2                              θ =   45° θ = 45°                                   Equation (36) F.sub.1 ' + F.sub.2 ' = F.sub.T = w = 1.156 ×             10.sup.7 K.sub.fρ  A.sub.j Sinθ(.increment.φ.sub.1 +            .increment.φ.sub.2)                                                       ΣF = 0 = F.sub.T - w, F.sub.T = w                                       1.156 × 10.sup.7 K.sub. fρ Sinθ A.sub.j (.increment.φ.    sub.1 + .increment.φ .sub.2) = .045864 lb.                                (1.156 × 10.sup.7) K.sub.f (5.253 × 10.sup.-7) (.707)             .00077(5 + 5.9) = .045864                                                     .036033 K.sub.f = .045864                                                     K.sub.f = 1.272829                                                            ______________________________________                                    

                  TABLE III                                                       ______________________________________                                        EXAMPLE 2: Calculation of K.sub.f                                             ______________________________________                                        w =         can weight = 13.9 Gm = .0306495 lbf                               φ.sub.1 =                                                                             2.6        φ.sub.2 =                                                                           3.8                                          A.sub.1 =   .00077 ft.sup.2                                                                          A.sub.2 = .00077 ft.sup.2                              θ.sub.1 =                                                                           45° θ.sub.2 =                                                                         45°                                   ΣF =  0                                                                 (36) F.sub.1 ' + F.sub.2 ' = F.sub.T = w                                      1.156 × 10.sup.7 K.sub.fρ SinθA.sub.j (.increment.φ.su    b.1 + .increment.φ.sub. 2) = .0306495 lbf.                                (1.156 × 10.sup.7) (5.253 × 10.sup.-7) (.707) (.00077) (2.6 +     3.8)K.sub.f =                                                                 .0306495                                                                      .02115706 K.sub.f = .0306495                                                  K.sub.f = 1.486653                                                             ##STR1##                                                                     ______________________________________                                    

                  TABLE IV                                                        ______________________________________                                        DETERMINATION OF K.sub.j                                                      Deadheaded - No velocity past Jets                                            Vent Diameter = 1"                                                                   Measured  Calc.                                                        Jet Diameter                                                                           φ   Pm      Pc    (Pc) × (K.sub.d)                                                                 K.sub.j                               ______________________________________                                        1/2      7       1.9     1.39  2.09     .909                                           6       1.65    1.19  1.79     .922                                           5       1.35    .997  1.49     .906                                           4       1.1     .797  1.19     .924                                           3       .85     .598  .897     .947                                           2       .55     .398  .598     .919                                           1       .28     .199  .299     .936                                   ##STR2##                                                                              K.sub.j Average = .923                                                ##STR3##                                                                     ______________________________________                                         φ & Pm are measured values of static pressure in supply duct and can      duct respectively                                                             Pc is the calculated value of Pm without empirical constants where            ##STR4##                                                                 

                  TABLE V                                                         ______________________________________                                        DETERMINATION OF K.sub.j                                                      Deadheaded - No velocity past Jets                                            Vent Diameter = 1"                                                                   Measured  Calc.                                                        Jet Diameter                                                                           φ   Pm      Pc    (Pc) × (K.sub.d)                                                                 K.sub.j                               ______________________________________                                        7/16"    7       1.35    .896  1.472    .917                                           6       1.15    .768  1.26     .912                                           5       1.0     .640  1.05     .952                                           4       .8      .512  .841     .951                                           3       .6      .384  .631     .950                                           2       .36     .256  .420     .857                                           1       .18     .128  .210     .857                                   ##STR5##                                                                              K.sub.j Average = .913                                                ##STR6##                                                                     ______________________________________                                    

                  TABLE VI                                                        ______________________________________                                        DETERMINATION OF K.sub.j                                                      Deadheaded - No velocity past Jets                                            Vent Diameter = 1"                                                                   Measured  Calc.                                                        Jet Diameter                                                                           φ   Pm      Pc    (Pc) × (K.sub.d)                                                                 K.sub.j                               ______________________________________                                        3/8"     8       1       .591  1.084    .922                                           7       .95     .518  .948     1.002                                          6       .85     .813  .813     1.04                                           5       .75     .677  .677     1.10                                           4       .55     .542  .542     1.01                                           3       .4      .406  .406     .985                                           2       .26     .271  .271     .959                                           1       .12      .1355                                                                              .1355    .88                                    ##STR7##                                                                              K.sub.j Average = .987                                                ##STR8##                                                                     ______________________________________                                    

                  TABLE VII                                                       ______________________________________                                        DETERMINATION OF K.sub.j                                                      Deadheaded - No velocity past Jets                                            Vent Diameter = 1"                                                                   Measured  Calc.                                                        Jet Diameter                                                                           φ   Pm      Pc    (Pc) × (K.sub.d)                                                                 K.sub.j                               ______________________________________                                        5/16"    8       .65     .295  .619     1.05                                           7       .55     .258  .542     1.01                                           6       .5      .221  .464     1.08                                           5       .4      .84   .387     1.03                                           4       .33     .147  .310     1.06                                           3       .25     .110  .232     1.07                                           2       .16      .0737                                                                              .1548    1.033                                          1       .07      .0369                                                                              .0773    .906                                   ##STR9##                                                                              K.sub.j Average = 1.03                                                ##STR10##                                                                    ______________________________________                                    

                  TABLE VIII                                                      ______________________________________                                        DETERMINATION OF K.sub.j                                                      Deadheaded - No velocity past Jets                                            Vent Diameter = 1"                                                                   Measured  Calc.                                                        Jet Diameter                                                                           φ   Pm      Pc    (Pc) × (K.sub.d)                                                                 K.sub.j                               ______________________________________                                        1/4"     8       .4      .139  .347     1.15                                           7       .36     .108  .269     1.33                                           6       .32     .0925 .231     1.38                                           5       .25     .0771 .193     1.30                                           4       .2      .0616 .154     1.29                                           3       .15     .0463 .116     1.29                                           2       .1      .0308 .077     1.29                                           1       .5      .0154 .0385    1.30                                   ##STR11##                                                                             K.sub.j Average = 1.29                                                ##STR12##                                                                    ______________________________________                                    

                  TABLE IX                                                        ______________________________________                                        DETERMINATION OF K.sub.j                                                      Deadheaded - No velocity past Jets                                            Vent Diameter = 1"                                                                   Measured  Calc.                                                        Jet Diameter                                                                           φ   Pm      Pc     (Pc) × (K.sub.d)                                                                K.sub.j                               ______________________________________                                        3/16"    8       .25     .0394  .124     2.01                                          7       .22     .0345  .109     2.01                                          6       .19     .0296  .0936    2.03                                          5       .15     .0246  .078     1.92                                          4       .12     .0197  .0624    1.92                                          3       .085    .0147  .0468    1.82                                          2       .055     .00985                                                                              .0312    1.76                                          1       .025    .0049  .015     1.67                                  ##STR13##                                                                             K.sub.j Average = 1.89                                                ##STR14##                                                                    ______________________________________                                    

                  TABLE X                                                         ______________________________________                                        DETERMINATION OF K.sub.v                                                      (Zero Velocity in Supply Duct)                                                Vent Diameter = 11/2" A.sub.v = .0122 ft.sup.2                                                      Calcu-        (Pc) ×                              Jet        Measured   lated   (Pc) ×                                                                        (K.sub.d) ×                         Diameter   φ  Pm      Pc    (K.sub.d)                                                                           (K.sub.j)                                                                           K.sub.v                           ______________________________________                                        1/2"       5      .42     .237  .473  .430  .977                              A.sub.j = .00136 ft.sup.2                                                                4      .325    .189  .378  .344  .945                              K.sub.j = .923 approx.                                                                   3      .24     .142  .284  .258  .930                              from graph FIG.                                                                          2      .16      .0947                                                                              .189  .172  .930                              11         1      .075     .0473                                                                               .0946                                                                              .860  .930                               ##STR15## K.sub.v Average = .942                                              ##STR16##                                                                    ______________________________________                                    

                  TABLE XI                                                        ______________________________________                                        DETERMINATION OF K.sub.v                                                      (Zero Velocity in Supply Duct)                                                Vent Diameter × 11/4" A.sub.v = .00852 ft.sup.2                                                             (Pc) ×                              Jet        Measured   Calc.   (Pc) ×                                                                        (k.sub.d) ×                         Diameter   φ   Pm     Pc    (K.sub.d)                                                                           (K.sub.j)                                                                           K.sub.v                           ______________________________________                                        1/2"       5       .7     .462  .809  .738  .948                              A.sub.j = .00136 ft.sup.2                                                                4       .55    .369  .647  .588  .935                                         3       .39    .277  .485  .441  .884                                         2       .26    .184  .323  .294  .884                                         1       .13    .09   .162  .147  .884                               ##STR17## K.sub.v Average = .907                                              ##STR18##                                                                    ______________________________________                                    

                  TABLE XII                                                       ______________________________________                                        DETERMINATION OF K.sub.v                                                      (Zero Velocity in Supply Duct)                                                Vent Diameter = 1" A.sub.v = .00545 ft.sup.2                                                                      (Pc) ×                              Jet        Measured  Calc.   (Pc) ×                                                                         (K.sub.d) ×                         Diameter   φ  Pm     Pc    (K.sub.d)                                                                            (K.sub.j)                                                                           K.sub.v                           ______________________________________                                        1/2"       5      1.4    .997  1.495  1.36  1.03                              A.sub.j × .00136 ft.sup.2                                                          4      1.1    .798  1.197  1.089 1.01                                         3      .8     .598  .897   .816   .980                                        2      .55    .398  .597   .542  1.01                                         1      .27    .99   .2985  .272   .992                              ##STR19## K.sub.v Average = 1.004                                             ##STR20##                                                                    ______________________________________                                    

                  TABLE XIII                                                      ______________________________________                                        DETERMINATION OF K.sub.v                                                      (Zero Velocity in Supply Duct)                                                Vent Diameter = 3/4" A.sub.v = .003068 ft.sup.2                                                                  (Pc) ×                               Jet        Measured  Calc.   (Pc) ×                                                                        (K.sub.d) ×                          Diameter   φ  Pm     Pc    (K.sub.d)                                                                           (K.sub.j)                                                                            K.sub.v                           ______________________________________                                        1/2"       5      2.6    2.20  2.75  2.50   1.04                              A.sub.j  = .00136 ft.sup.2                                                               4      2.05   1.76  2.20  2.002  1.02                                         3      1.5    1.32  1.65  1.50   1.0                                          2      1.0    .880  1.10  1.001  1.0                                          1      .5     .440  .550  .5005  1.0                                ##STR21## K.sub.v Average = 1.012                                             ##STR22##                                                                    ______________________________________                                    

                  TABLE XIV                                                       ______________________________________                                        DETERMINATION OF K.sub.v                                                      (Zero Velocity in Supply Duct)                                                Vent Diameter = 1/2 A.sub.v = .00136 ft.sup.2                                                                    (Pc) ×                               Jet        Measured  Calc.   (Pc) ×                                                                        (K.sub.d) ×                          Diameter   φ  P      P     (K.sub.d)                                                                           (K.sub.v)                                                                            K.sub.v                           ______________________________________                                        1/2"       5      4.1    4.0   4.0   3.64  1.127                              A.sub.j = .00136 ft.sup.2                                                                4      3.25   3.2   3.2   2.91  1.02                                          3      2.4    2.4   2.4   2.184 1.10                                          2      1.6    1.6   1.6   1.45  1.10                                          1      .80    .80   .80   .728  1.10                                ##STR23## K.sub.v Average = 1.11                                              ##STR24##                                                                    ______________________________________                                    

Using the above relationships, a design of an elevator in accordancewith the present invention is hereinafter illustrated.

A. Basic Requirements

1. 1000 cans per minute

2. 6" spacing on centerline of cans

3. Elevator height 10 ft.-Discharge 1.6 ft.

4. Can duct static pressure at inlet=0.4 in. H₂ O maximum

B. Basic Assumptions

1. V_(j) Sin θ=7V at inlet

2. Jet on 11/2" centers

3. Vents on 11/2" centers

4. 3 jets act on each can end to give lifting force on can

5. Assume θ at inlet=45°, and maximum θ≦60°

C. Calculation of force required to maintain velocity (Assume cans aremoving on frictionless horizontal plane) ##EQU12## where weight of 1can=0.03116 lbf ##EQU13##

    F=0.13445 lbf

This is force required to maintain velocity. This force must be impartedto the cans through all jets acting on the cans.

D. Calculation of number of cans in the elevator and discharge at onetime. ##EQU14## E. Calculation of force exerted on the column due to thestatic pressure in can duct. If P=0.4 in. H₂ O and 1 in. H₂ O=5.328lbf/ft² and cross sectional area of a #209-12 oz. can is 0.09115 ft²,the force exerted is: ##EQU15##

    F=0.19425 lbf

F. Calculation of weight of all cans in elevator at one time. Each canweighs 0.03116 lbf so the total weight of the column is: ##EQU16## G.Calculation of Force that must be supplied through all jets to supportall cans in the can duct.

    Cans weight=0.7229 lbf.

Force due to static pressure in can duct=0.19425 lbf. Therefore, theforce the jets must supply to support that portion of weight not alreadysupported by the static pressure between the cans is:

    F=0.7229-0.19425=0.5287 lbf

H. Calculation of number of jets acting on the cans. Since six jet totalact on each can: ##EQU17## I. Calculation of effective weight of cans.With 23.2 cans supported by a force of 0.19425 lbf, the effective weightof each can is:

    0.19425 lbf/23.2 cans=0.0083728 lbf/can

J. Calculation of total force required through the jets. Force tomaintain velocity is=0.13445 lbf distributed among the 139.2 jets actingon the cans, this is

    0.13445 lbf/139.2 jets=0.0009659 lbf/jet

Force to lift required weight is=0.5287 lbf distributed among the 139.2jets is 0.5287 lbf/139.2 jets=0.003797 lbf/jet, so the total force thrueach jet is 0.003797+0.0009659=0.004763 lbf/jet

K. Calculation of inlet conditions.

1. Assume V_(j) Sin θ=7V at inlet, V=500 ft/min so V_(j) Sin θ=7(500ft/min)=3500 ft/min ##EQU18## Δφ to give this velocity is V_(j) =3400√Δφ##EQU19##

    φ=P+Δφ=0.4+2.12=2.52 in. H.sub.2 O static pressure

(K) F=K_(f) ρA_(j) (V_(j) Sin θ-V)², where K_(f) =1.36, reference TableIII ##EQU20## but K_(d) and K_(v) are both functions of A_(v), so thesolution is most easily found by trial and error as follows:

Trial 1: ##EQU21## Since A_(v) =0.4876 in.², D_(v) =0.788 in. diameterTrial 2: ##EQU22## From Graph FIG. 11, K_(j) =0.96 and from FIG. 12,K_(v) =1.0 ##EQU23## A_(v) =0.617 in.², D_(v) =0.887 in. diameter Trial3: ##EQU24## A_(v) =0.647 in.², D_(v) =0.907 in. diameter Trial 4:##EQU25## A_(v) =0.6527 in.², D_(v) =0.912 in. diameter, and since thisdiameter at Trial 4 varies only 0.005 with the diameter at Trial 3, itcan be considered sufficiently accurate to end the trial and errorsolution here.

L. Calculation of conditions at 2 ft. At 2 ft. there are 9.6 ft. of cancolumn above so there are ##EQU26## The effective weight of this columnis: ##EQU27## With a cross-sectional area of 0.09115 ft², the can ductstatic pressure is: ##EQU28## so P=0.331 in. H₂ O at 2 ft. Since φ=2.52in. H₂ O, Δφ=φ-P=2.52-0.331=2.189 So V_(j) =3400√2.189=5030.4 ft/min##EQU29## Trial 1: ##EQU30## A_(j) =0.1269 in.². D_(j) =0.402 in. Trial2: At D_(j) =0.402 in. and D_(v) =0.912 in. ##EQU31## From Graphs FIGS.11 and 12, K_(j) =0.95, K_(v) =0.99. ##EQU32## A_(j) =0.0999 in.², D_(j)=0.357 in. diameter. (L) Trial 3: ##EQU33## K_(j) =(0.97) and K_(v)=(0.99) from FIGS. 11 and 12 ##EQU34## A_(j) =0.0953 in.², D_(j) =0.348in. diameter. Since this varies only 0.009 in. with the preceding valueof D_(j), we will assume this value to be sufficiently accurate.

    F=K.sub.f ρA.sub.j (V.sub.j Sin θ-V).sup.2 ##EQU35##

    θ=47.9°

M. Calculation of conditions at 4 ft. At 4 ft. there are 7.6 ft. ofcolumn above so there are: ##EQU36## The effective weight of the columnis 15.2×0.0083728 lbf=1.27267 lbf. With cross-sectional area=0.09115ft.², the can duct static pressure is:

    0.127267 lbf/0.09115 ft..sup.2 =1.3962 lbf/ft..sup.2 ##EQU37## So P=0.262 in. H.sub.2 O at the 4 ft. elevation. Since φ=2.52, Δφ=φ-P=2.52-0.262=2.258 in. H.sub.2 O,

    V.sub.j 3400√Δφ=3400√2.258=5109.1 ft/min

Since D_(v) =0.912 in., A_(v) =0.6527 in.² ##EQU38## (M) Trial 1:##EQU39## A_(j) =0.1111 in.², D_(j) =0.376 in. diameter Trial 2:##EQU40## From Graphs FIGS. 11 and 12, K_(j) =(0.96) and K_(v) =(1)approximately: ##EQU41## A_(j) =0.084756 in.², D_(j) =0.3285 in.diameter Trial 3: ##EQU42## From Graphs FIGS. 11 and 12, K_(j) =1.02 andK_(v) =1.0 ##EQU43## A_(j) =0.0779 in.², D_(j) =0.315 in. diameter.Since this is only 0.014 from the preceding Trial, one can consider thisvalue sufficiently accurate.

    F=K.sub.f ρA.sub.j (V.sub.j Sin θ-V).sup.2 ##EQU44##

    θ=51.7°

N. Calculation of conditions at 6 ft. At 6 ft. there are 5.6 ft. ofcolumn above so there are: ##EQU45## The effective weight of this columnis 11.2×0.0083778 lbf=0.09377536 lbf. With cross-sectional area=0.09115ft.², the can duct static pressure is: ##EQU46## so

    P=0.1931 in. H.sub.2 O

    Δφ=φ-P=2.52-0.1931=2.3269 in. H.sub.2 O

    V.sub.j =3400√Δφ=3400√2.3269=5186.4 ft/min ##EQU47## Trial 1: ##EQU48##  D.sub.j =0.346 in. Trial 2: ##EQU49##  From Graphs FIGS. 11 and 12, K.sub.j =(0.102), K.sub.v =(1.0) approx. ##EQU50##  D.sub.j =0.294 in. Trial 3: ##EQU51##  From Graphs FIGS. 11 and 12, K.sub.j =(1.05), K.sub.v =(1.0) ##EQU52##  A.sub.j =0.06268 in..sup.2, D.sub.j =0.283 in. Since this varies only 0.011 with the last value one can consider this sufficiently accurate.

    F=K.sub.f ρA.sub.j (V.sub.j Sin θ-V).sup.2 ##EQU53##  θ=58.3°

(N) Due to design considerations, jet angles above 60° are undesirable.Therefore, in the calculations of conditions at 8 ft. and above, the jetdiameter and angle will remain at approximately 0.283 inches (adjustingonly for the changes in Δφ) and 58.3°, to give the correct force neededto maintain the can velocity. The outlet vent diameter will increase togive the proper pressure ratio between the supply duct and can duct.

O. Calculation of conditions at 8 ft. At 8 ft. there are 3.6 ft. ofcolumn above so there are: ##EQU54## The effective weight of these cansis 7.2 cans×0.00837728 lbf=0.06031584 lbf. With a cross-sectional areaof 0.09115 ft² the can duct static pressure is: ##EQU55## so P=0.124 in.H₂ O, Δφ=φ-P=2.52-0.124=2.396 in. H₂ O ##EQU56##

    F=K.sub.f ρA.sub.j (V.sub.j Sin θ-V).sup.2  A.sub.j =0.000421 ft.sup.2 =0.060624 in..sup.2, D.sub.j =0.278 in. ##EQU57## (O) Trial 1: ##EQU58##  D.sub.v =0.8234 in. diameter Trial 2: ##EQU59##  From Graphs FIGS. 11 and 12, K.sub.j =(1.15), K.sub.v =(1.0) approx. ##EQU60##  A.sub.v =0.816 in..sup.2, D.sub.v =1.02 in. Trial 3: ##EQU61##  From Graphs FIGS. 11 and 12, K.sub.j =(1.15), K.sub.v =(0.98) ##EQU62##  A.sub.v =0.87806 in..sup.2, D.sub.v =1.05 in. Trial 4: ##EQU63##  From Graphs FIGS. 11 and 12, K.sub.j =1.15, K.sub.v =(0.98) ##EQU64##  A.sub.v =0.891 in..sup.2, D.sub.v =1.065 in. Since this diameter is within 0.015 of the last Trial, it can be considered sufficiently accurate.

P. Calculation of conditions at 10 ft. At 10 ft. there are 1.6 ft. ofequivalent height left in discharge elbow. At 0.5 ft. per can there are##EQU65## Since the effective weight of the can is 0.0883728 lbf, theeffective column weight is: (3.2) (0.0083728)=0.02679 lbf Withcross-sectional area of can=0.09115 ft², this will require a can ductstatic pressure of: ##EQU66## so P=0.05517 in. H₂ O,Δφ=φ-P=2.52-0.05517=2.46 in. H₂ O ##EQU67##

    F=K.sub.f ρA.sub.j (V.sub.j Sin θ-V).sup.2  A.sub.j =0.000408 ft.sup.2 =0.05877 in..sup.2, D.sub.j =0.273 in. ##EQU68## Trial 1: ##EQU69##  A.sub.v =0.78564 in..sup.2, D.sub.v =1.000 in. Trial 2: ##EQU70##  From Graphs FIGS. 11 and 12, K.sub.j =(1.2), K.sub.v =(0.99) ##EQU71##  D.sub.v =1.29 in. Trial 3: ##EQU72##  From Graphs FIGS. 11 and 12, K.sub.j =(1.2), K.sub.v =(0.94) ##EQU73##  A.sub.v =1.425 in..sup.2, D.sub.v =1.347 in. Trial 4: ##EQU74##  From Graphs FIGS. 11 and 12, K.sub.j =(1.2), K.sub.v =(0.94) ##EQU75##  A.sub.v =1.44 in..sup.2, D.sub.v =1.358 in. (P) Since this varies only 0.009 with the previous Trial diameter, it can be considered sufficiently accurate.

Therefore, one solution to the elevator design fulfilling the basicrequirements described above is set forth on TABLE XV as follows:

                  TABLE XV                                                        ______________________________________                                        ELEVATOR DESIGN PARAMETERS                                                            P        φ   D.sub.j                                                                            D.sub.v                                                 (In.     (In.    (in- (in- θ                                                                              Velocity                            Elevation                                                                             H.sub.2 O)                                                                             H.sub.2 O)                                                                            ches)                                                                              ches)                                                                              (degrees)                                                                            Ratio                               ______________________________________                                        10 ft    .05517  2.52    .273 1.358                                                                              58.3°                                                                         9.1 to 1                            8 ft.   .124     2.52    .278 1.065                                                                              58.3°                                                                         8.96 to 1                           6 ft.    .1931   2.52    .283  .912                                                                              58.3°                                                                         8.83 to 1                           4 ft.   .262     2.52    .315  .912                                                                              51.7°                                                                         8.02 to 1                           2 ft.   .331     2.52    .348  .912                                                                              47.9°                                                                         7.35 to 1                           Inlet   .400     2.52    .367  .912                                                                              45°                                                                           7 to 1                              ______________________________________                                    

The spacing of the cans is established at the feed or inlet end of theelevator and may suitably be established by the in-feed of thisinvention. The air elevator in and of itself has no provision forestablishing the spacing between cans. This function is provided by thein-feeder generally identified by reference numeral 60 and roughlyextending from point A to point B. The in-feeder 60 is usually a slightbit more than a 90° bend but need not be. The can duct 18a of thein-feeder is very similar to the can duct 18 of the elevator with theexception of the absence of multiple exhaust openings. Air is introducedinto the can duct 18a through jets 51a and 52a from the air supply duct24a which may be an extension of duct 24 or be separate and providedwith a separate fan or blower. A sliding damper 62 is provided on theinside side or enclosure 30a just before and adjacent the juncture ofthe in-feed 60 with the vertical elevator 10 at point B. The slidingdamper 62 serves to open or close an opening 64 in the side or enclosure30a. The cans 12a coming into the inlet 66 of the in-feeder 60 areeither under mechanical, gravity or air pressure as where the jet boards54a and 55a extend to the right in FIG. 2 beyond the can duct enclosures28a and 30a. A force must be applied to the cans to move them into thein-feeder since the air being introduced into the in-feeder can duct 18atends to produce a back pressure therein and acts to push to the rightin FIG. 2 and prevent cans 12a from entering the in-feeder can duct 18a.As cans 12a are pushed into the in-feeder can duct 18a against the airpressure, the cans 12a will progress up the in-feed to a point 66 justprior to the opening 64 and as a can passes this point, the air pressurebetween this can and the previous can will propel the can at point 66forward, with the air ahead of this can escaping through opening 64 thusproviding a spacing 67 between cans entering the elevator 10 the size ofwhich depends upon the positioning of damper 62. The greater the opening64, the greater the volume of air that is allowed to escape throughopening 64 by the positioning of damper 62, and the closer the cans arespaced for introduction into the elevator 10. The volume of air, CFM,injected into the elevator 10 and the static pressure produced are suchas to maintain the selected spacing of the cans 12.

At the output 68 of the elevator 10 at point C, it is usually desirableto feed the elevated cans at a slower rate and more nearly in can-to-cancontact to a work station 70 and the like. To accomplish thistransition, there may be provided a de-accelerator 72 having an inlet 74connected to receive cans from the outlet 68 of elevator 10, and anoutlet 76. The de-accelerator 72 comprises a can duct 18b made up ofends 20b and 22b each having jets 51b and 52b therein or in jet boards54b and 55b similar to that previously explained with regard to elevator10. The ends 20b and 22b may form a common wall between the can duct 18band air supply ducts 24b and 26b. For at least the central portion ofthe de-accelerator 72, there are provided a greater number of jets 51band 52b which are considerably closer together than those of theelevator 10 (see FIG. 5). This arrangement of jet holes permits a muchhigher volume of air to be introduced in the active portion of thede-accelerator and provides a barrier or cushion of air such that thecans 12b are permitted to close upon one another to effectively providea braking effect on the cans 12 exiting the outlet 68 of elevator priorto their reaching the outlet 76 of the de-accelerator 72 and arriving atthe work station 70. It will be noted that the jet holes 51b and 52b areessentially twice in number and thus introduce twice the volume of aircompared to those of the elevator to provide an increased amount of airintroduced within the active area. The active area is defined as about15° to either side of the bisector of the included angle of discharge.The maximum bisector angle is no greater than 45° although the dischargeangle may exceed 90°. The angle of the jets in this portion may bereduced also to reduce the forward speed of the cans.

The conveyor 73, shown dashed in FIG. 1, and as set forth in FIG. 7,comprises a different embodiment known as a fan wherein the curvature ofthe can duct is at 90° with respect to the can duct of thede-accelerator 72 such that the cans 12c are moved in a fan pattern suchthat the axis 78 of the can is rotated about a point 80 along its lengthto present the cans 12c in an upright position to the work station 70cwhich may be a horizontal elevator or transfer unit 82. Since the airjets in the fan, once the cans have moved through about 45°, will haveless gravity to overcome, the jets beyond about 45° may be fewer innumber and/or smaller. It is important, however, that the cans bemaintained in a proper position such that the trailing edge of the openend contacts the outer radius of the fan to avoid excessive elongationof the can. The jets are positioned and adjusted to slightly tip theopen end 14d of the cans in the direction of travel to avoid contact ofthe leading edge with the can duct. This is the case also where the openend of the can is positioned downwardly.

Referring to FIG. 7, the transfer unit 82 is quite similar inconstruction to the vertical elevator 10 in that there is provided a canduct 18d positioned to transport the can 12d with its axis of rotationpositioned vertically. The can duct 18d is formed with ends 20d and 22dprovided with jets 52d and 53d therein or in jet boards 54d and 55d. Airsupply ducts 24d and 26d are connected to suitable air supplies whichsupply air to the ducts and via jets 52d and 53d to the can duct 18d.The sides 28d and 30d may each be provided with a series of spacedexhaust holes 31d and 33d to provide and maintain a positive staticpressure with respect to the ambient atmosphere between articles in thearticle conveyor duct. Again, the size and positioning of the jetsshould be such that the open end of cans 12d are tilted slightly in thedirection of travel.

In an embodiment of FIGS. 1 and 2 wherein the jets 51 and 52 of theelevator 10 are of the same size throughout the height thereof, thespacing established between cans 12 by the in-feed 60 is maintained byintroducing air into the space 22 and exhausting the air through exhaustopenings 31 such as to maintain a positive static pressure within thespace between the cans. Where the amount of air introduced in oneportion of the elevator is greater or less than that introduced intoanother, the spacing between the cans may be varied as the cans traversethe length of the elevator. For example, if the size of the jet openingswere increased, at the same supply static pressure, in relation toheight of the elevator, the greater amount of air introduced betweencans would tend to cause the cans to separate still further as theyprogress up the elevator. Conversely, if the jets were reduced in size,with the same supply static pressure, the spacing between the cans woulddiminish as they progress up the elevator. The same phenomenon would beobserved where the jet sizes were held constant and the supply staticpressure were increased or decreased with respect to the height of theelevator.

Referring now to FIG. 6, there is shown in cross section to an enlargedscale, a length of the jet board 152a having a series of apertures 154therein adapted to receive a nozzle 156 having a spherical ball 158permitting angular movement of the nozzle within the jet board. Anactuator board 160 is slidably attached to the jet board 152a andreceives through a series of apertures therein one extremity of each ofthe nozzles 156 such that on sliding movement of the actuator 160 all ofthe nozzles are movable. Such an arrangement provides for selectivelyadjusting the angle of nozzles 156 as desired or automatically as bybeing under control of a computer and the like.

What is claimed is:
 1. An articles transportation system fortransporting a series of articles to provide a predetermined spacingtherebetween comprising in combination:an elongated article conveyorduct with an inlet and an outlet having side walls and end walls spacedapart a distance just sufficient to accommodate an article within theduct with a predetermined clearance between the article and the side andend walls; a pair of guide rails positioned along the interior of theside walls to support the articles in predetermined spaced relation tothe side walls; a series of spaced downstream slanted jet openings alongthe length of the end wall; a plenum chamber forming an air supply ductextending along the end walls in communication with said jet openings;means for supplying air under pressure to the plenum chamber forsupplying jets of air via said jet openings to the article duct in adownstream direction; exhaust opening means serially positioned along atleast one of said side walls being of a size in relation to the jetopenings and the amount of air introduced into the article duct via thejet openings to provide and maintain a positive static pressure withrespect to the ambient atmosphere between articles in the articleconveyor duct; and feeder means, for the articles to be transported,connected to the inlet of the article conveyor duct for feeding articlesthereto at a selected uniform rate at spaced intervals, said feedermeans comprising a length of duct connected to the inlet of the articleconveyor duct having side walls and end walls with a series of spaceddownstream slanted jet openings along the length of the end walls and asingle exhaust opening in a side wall located adjacent the inlet end ofthe article conveyor duct.
 2. The combination of claim 1 wherein theexhaust opening means in the article conveyor duct side comprise aplurality of circular openings serially positioned along the length ofthe container duct.
 3. The combination of claim 1 wherein the exhaustopening means comprise an elongated slot in the side of the articleconveyor duct along the length thereof.
 4. The combination of claim 1wherein the exhaust opening means is comprised of a series of slotspositioned laterally across the width of a side.
 5. The combination ofclaim 1 wherein the exhaust opening is of rectangular configuration andis provided with an adjustable shutter for selective adjustment of theeffective size of the exhaust opening.
 6. The combination of claim 1including de-accelerator means connected to the outlet end of thearticle duct to de-accelerate the speed of the articles issuingtherefrom and reduce the predetermined spacing therebetween.
 7. Thecombination of claim 6 wherein the de-accelerator means comprises alength of duct connected to the outlet of the article duct having sidewalls and end walls with a series of spaced jet openings along thelength of the end walls and exhaust means serially positioned along atleast one of the side walls, wherein the jet openings are approximatelytwice in number per unit of length of duct as the jet openings in thearticle duct.
 8. The system of claim 1 wherein the air jet openings areadapted to introduce air into the article conveyor duct at an angle offrom about 18 degrees to about 60 degrees with respect to a line normalto the longitudinal axis of the article conveyor duct.
 9. Thetransportation system according to claim 1 wherein the air volumesupplied to the inlet of the supply duct in relation to the total crosssectional area of the air jet openings along with the static pressure tobe maintained across the jets and the cross sectional area of the supplyduct are selected to provide a substantially uniform static pressure inthe air supply duct throughout its length.
 10. The transportation systemaccording to claim 1 wherein the volume of air supplied to the inlet ofthe supply duct divided by the total cross sectional area of the jetopenings is substantially equal to the square root of the staticpressure difference across the length of the jet openings times thevelocity of the air through the jets at 1 inch H₂ O static pressuredifferential across the length of the jet openings and the supply ductvelocity is maintained to provide a substantially constant staticpressure throughout the length of the supply duct.
 11. Thetransportation system of claim 1 wherein the exhaust opening and jetopenings sizes are selected with respect to the static pressuredifferential between the air supply duct and the article duct to providea positive static pressure with respect to the ambient atmospherebetween articles in the conduit substantially equal to ##EQU76## whereA_(j) is the cross sectional area of the jet, A_(v) is the crosssectional area of the exhaust opening, φ is the static pressure in thesupply duct, N is the number of jets per vent and K_(d), K_(j) and K_(v)are empirical modifiers.
 12. The transportation system of claim 11 whichincludes a force per jet acting on the article substantially equal to

    F=K.sub.f ρA.sub.j (V.sub.j Sin θ-V).sup.2

where ρ is the density of the air, V_(j) is the velocity of air throughthe jet, A_(j) is the cross sectional area of the jet, θ=angle of jet,K_(f) is an empirical constant and V is the velocity of the article.